>>> str(datetime.timedelta(hours=10.56))
10:33:36

>>> td = datetime.timedelta(hours=10.505) # any timedelta object
>>> ':'.join(str(td).split(':')[:2])
10:30

将timedelta对象传递给str()函数调用的格式化代码与输入print td相同。因为你不想要秒，我们可以用冒号分隔字符串(3部分)，然后只用前2部分把它重新组合在一起。

我的datetime。Timedelta对象的时间大于一天。这是一个更深层次的问题。以上所有讨论都假设时间不超过一天。timedelta实际上是一个由日、秒和微秒组成的元组。上面的讨论应该使用td。秒，因为乔做了，但如果你有天，它不包括在秒值。

我得到2个日期时间和打印天和小时之间的时间跨度。

span = currentdt - previousdt
print '%d,%d\n' % (span.days,span.seconds/3600)

按照上面Joe的示例值，我将使用模算术运算符，因此:

td = datetime.timedelta(hours=10.56)
td_str = "%d:%d" % (td.seconds/3600, td.seconds%3600/60)

注意，Python中的整数除法默认是四舍五入;如果想要更显式，可以适当使用math.floor()或math.ceil()。

def seconds_to_time_left_string(total_seconds):
    s = int(total_seconds)
    years = s // 31104000
    if years > 1:
        return '%d years' % years
    s = s - (years * 31104000)
    months = s // 2592000
    if years == 1:
        r = 'one year'
        if months > 0:
            r += ' and %d months' % months
        return r
    if months > 1:
        return '%d months' % months
    s = s - (months * 2592000)
    days = s // 86400
    if months == 1:
        r = 'one month'
        if days > 0:
            r += ' and %d days' % days
        return r
    if days > 1:
        return '%d days' % days
    s = s - (days * 86400)
    hours = s // 3600
    if days == 1:
        r = 'one day'
        if hours > 0:
            r += ' and %d hours' % hours
        return r 
    s = s - (hours * 3600)
    minutes = s // 60
    seconds = s - (minutes * 60)
    if hours >= 6:
        return '%d hours' % hours
    if hours >= 1:
        r = '%d hours' % hours
        if hours == 1:
            r = 'one hour'
        if minutes > 0:
            r += ' and %d minutes' % minutes
        return r
    if minutes == 1:
        r = 'one minute'
        if seconds > 0:
            r += ' and %d seconds' % seconds
        return r
    if minutes == 0:
        return '%d seconds' % seconds
    if seconds == 0:
        return '%d minutes' % minutes
    return '%d minutes and %d seconds' % (minutes, seconds)

for i in range(10):
    print pow(8, i), seconds_to_time_left_string(pow(8, i))


Output:
1 1 seconds
8 8 seconds
64 one minute and 4 seconds
512 8 minutes and 32 seconds
4096 one hour and 8 minutes
32768 9 hours
262144 3 days
2097152 24 days
16777216 6 months
134217728 4 years

>>> str(datetime.timedelta(hours=10.56))
10:33:36

>>> td = datetime.timedelta(hours=10.505) # any timedelta object
>>> ':'.join(str(td).split(':')[:2])
10:30

将timedelta对象传递给str()函数调用的格式化代码与输入print td相同。因为你不想要秒，我们可以用冒号分隔字符串(3部分)，然后只用前2部分把它重新组合在一起。

下面是一个stringify timedelta.total_seconds()的函数。它在python2和python3中工作。

def strf_interval(seconds):
    days, remainder = divmod(seconds, 86400)
    hours, remainder = divmod(remainder, 3600)
    minutes, seconds = divmod(remainder, 60)
    return '{} {} {} {}'.format(
            "" if int(days) == 0 else str(int(days)) + ' days',
            "" if int(hours) == 0 else str(int(hours)) + ' hours',
            "" if int(minutes) == 0 else str(int(minutes))  + ' mins',
            "" if int(seconds) == 0 else str(int(seconds))  + ' secs'
        )

示例输出:

>>> print(strf_interval(1))
   1 secs
>>> print(strf_interval(100))
  1 mins 40 secs
>>> print(strf_interval(1000))
  16 mins 40 secs
>>> print(strf_interval(10000))
 2 hours 46 mins 40 secs
>>> print(strf_interval(100000))
1 days 3 hours 46 mins 40 secs